Let $S$ be a subset of $Ord$ where $Ord$ denotes the ordinal class. I aim to show that $Ord\setminus S$ is order-isomorphic to $Ord$.
I have spent some time on this question. The question seems not easy enough for me and I do not even know how to get start. So could someone please help? Thanks a lot!
EDIT:Now I notice that it is quite possible for the question to be related to the theorem:
And again I am stuck...I am wondering how I can find the relation between some class and some initial segement to apply the theorem. And to disprove one of (i) and (ii), maybe I need to find a contradiction. But so far I cannot see what is the contradiction here I am looking for.
So I am asking for help. Could someone please give some explicit explaination? Thanks so much!
EDIT: I am wondering why it is voted to be closed. If someone who can give some thoughts on this question provided some more details are added, I am always glad to edit my post. I am a beginner of MK set theory who appreciate rather detailed answer. Thanks for patient.
(1). If $T$ is a set of ordinals then there cannot be an injection $f:Ord$ \ $T \to U$ for any set $U$.
Else, $V= \{u\in U:\exists a\in Ord$ \ $T\; (u=f(a))\}$ is a set, so $\{f^{-1}(v):v\in V\} \cup T=Ord$ is a set. (But then $Ord +1=Ord \cup \{Ord\}$ is an ordinal that's larger than any ordinal.)
(2). Let $A=Ord$ and $B=Ord$ \ $S$ where $S\subset Ord$ is a set.
(i). Suppose $h:A\to B$ where $h$ is an order-isomorphic embedding of $A$ onto an initial segment of $B.$ Then by (1),(with $T=\phi$), we know that $h(A)$ cannot be a proper initial segment of $B$. That is, $\forall b\in B\;(h(A)\ne \{c\in B:c<b\}.$ So $h(A)=B.$
(ii). Suppose $g:B\to A$ is an order-isomorphic embedding of $B$ onto an initial segment of $A.$ Then by (1),(with $T=S$), we know that $g(B)$ cannot be a proper initial segment of $A.$ So $g(B)=A.$
(3). Another method would be to take the transitive (Mostowki) collapse of $B=Ord$ \ $S$. That is, for $b\in B$ let $t(b)=\{ t(c): b>c\in B\}.$ (E.g. $t(\min B)=0$ and $t(\min (B$ \ $\{\min B\})=1$).
Use transfinite induction on $B$ to show that $t(B)\subset Ord$ and that $t$ is an order-isomorphic embedding. Then use transfinite induction on $Ord$ to show that $t(B)\supset Ord.$